Question: Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $x \neq 0$. $q = \dfrac{-10x + 20}{x - 4} \times \dfrac{x^2 - 13x + 36}{6x - 12} $
Explanation: First factor the quadratic. $q = \dfrac{-10x + 20}{x - 4} \times \dfrac{(x - 4)(x - 9)}{6x - 12} $ Then factor out any other terms. $q = \dfrac{-10(x - 2)}{x - 4} \times \dfrac{(x - 4)(x - 9)}{6(x - 2)} $ Then multiply the two numerators and multiply the two denominators. $q = \dfrac{ -10(x - 2) \times (x - 4)(x - 9) } { (x - 4) \times 6(x - 2) } $ $q = \dfrac{ -10(x - 2)(x - 4)(x - 9)}{ 6(x - 4)(x - 2)} $ Notice that $(x - 2)$ and $(x - 4)$ appear in both the numerator and denominator so we can cancel them. $q = \dfrac{ -10\cancel{(x - 2)}(x - 4)(x - 9)}{ 6\cancel{(x - 4)}(x - 2)} $ We are dividing by $x - 4$ , so $x - 4 \neq 0$ Therefore, $x \neq 4$ $q = \dfrac{ -10\cancel{(x - 2)}\cancel{(x - 4)}(x - 9)}{ 6\cancel{(x - 4)}\cancel{(x - 2)}} $ We are dividing by $x - 2$ , so $x - 2 \neq 0$ Therefore, $x \neq 2$ $q = \dfrac{-10(x - 9)}{6} $ $q = \dfrac{-5(x - 9)}{3} ; \space x \neq 4 ; \space x \neq 2 $